Thursday, August 30, 2007

Pirates and counting

You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it?

20 comments:

SO he only needs two pirates to agree with the way he has divided the coins, so he only needs to give two of the oher 4 men coins and would want to give them the same amount (again so they do not disagree). The smallest number that can be divided between the two pirates is obviously 2. So one coin for each of the two and 98 for him (maximizing his wealth). However, in theory couldn't the two pirates who each got one coin feel that they deserved more and then disagree? Then the head pirate would be killed and the remaining pirates could potentially split the 100 coins 4 ways giving each man 25 coins.

The top pirate doesn't *need* to end up with 98percent of the gold, he just needs to not get killed. So he divides the gold into thirds (yes, they used to cut coins to make smaller denominations), giving one third to himself and the remaining two thirds to his two top pirates, and the three of them have the vote. That is, if the top pirate gets to vote. If he can only propose the division but not vote, his problem is a tad deeper, he needs to divide the gold into fourths... On the other hand, he's a pirate; just shoot the other 4 and take all the gold... ;-)

I believe that the only way that the first pirate to live and have any gold for himself is to take one coin for himself give 1 to the #2 guy and give the remaining 98 to the last prate. He would definitely get the last pirates vote. If pirate 2 rationalized that if he voted against this distribution that it would ultimately result in his demise I think the second pirate would vote to accept the distribution. This makes the assumption that if they killed one pirate that the next in line would be in charge of the distribution. It also assumes that the last pirate would only not kill if it made financial sense. What do you think?

Assuming that the pirates are motivated primarily by survival, then to a lesser extent by greed and finally to the least extent by sadism (i.e. they'd prefer to receive a gold coin and see someone get executed than just receive one coin earlier, but would prefer one coin to none and an execution; and obviously would prefer 0 coins and surviving to 100 coins and being executed), and act in a logical way, what is the maximum number of coins the eldest pirate can get?

Although the problem revolves around the eldest pirate, in order to solve the problem it is preferable to work from the youngest and work up to the case of eldest. Thus let the youngest pirate be A and the eldest pirate E.

A knows that if only he and B survive the earlier rounds, then B will get all 100 coins as his vote in favour counts as a majority. He also knows that B will prefer this outcome above all others and so will vote against any proposal made by C - even if C offers all 100 coins to B, he'd rather see C get executed and then take the 100 coins than just take the coins from C. As C will obviously vote for his own proposal, A therefore will vote in favour of any proposal by C which offers him just a single gold coin, as by voting for he receives a coin whereas by voting against he receives 0.

In order for his proposal to be accepted, D needs one other to vote with him. By offering 100 coins to C he can secure his vote, as this is more than the 99 he will receive if D is executed and he is forced to buy A's vote. However, B's vote can effectively be bought for 1 gold coin as he knows that if D is executed, A and C will split the coins between them and so he receives nothing. An offer of 0 to B is insufficient to secure his vote as he'd prefer to see D get executed and then receive 0 from C than simply accept 0 at this stage.

Thus in the case of 4 pirates, D receives 99 coins and B receives 1, whilst A and C receive 0. In order to get his proposal accepted E (the eldest in this case) requires the vote of 2 others, and it will obviously be easiest to get these votes from A and C who know they will get 0 coins if they vote against. In order to over-rule the sadism motive for A and C he only needs to offer them a single coin each.

Hence the eldest pirate can survive and keep 98 of the gold coins by offering 1 to A and C.

ANS: pirate one must either give it all away or be killed (why? because pirate 2 and 3 will hold out only for 50 and 50). But if he is killed, the dividing of the coins is the same process for the next highest ranked pirate, and the new top pirate faces the same dilemma, give it all away or be killed, until there are two pirates left, and the lower pirate disagrees to any offer except all of the coins. SO, pirate one explains this to the other pirates, who now understand, and pirates 2 and 3 know that they have to accept his minimum offer of 1 coin each or they get none. The top pirate keeps all 98 of the rest.

Pirate 4 knows that he will get killed and not keep any gold if he gets to propose how it is divided, since #5 would always vote no, so he is the most motivated to accept 1 coin. Pirate 3 knows that pirate 4 will accept 1 coin when #3 gets to propose the division, so pirate 3 will not vote for any division by pirate 1 because he wants to hold out for 99 coins. Pirate 2 also knows that pirate 4 would take 1 coin from him to vote for the division, but pirates 3 and 5 would never vote for the division by pirate 2. So Pirate 1 gives 1 coin to pirate 2, 1 coin to pirate 4, and keeps 98 coins for himself.

i blog offers a more nuanced, and developed answer, determining which of the two would most likely accept one coin each from pirate one. But the assumption is that all the pirates have an equal ability to reason out the logical conclusion, pirate three no less than pirate four; meaning that pirate three would as easily accept 1 coin knowing his other option is zero - seeing how favorable pirate four looks on.

final thought, i think prate 4 and 5 are the most likely to accept one coin each from pirate 1. 4 is obvious, but why 5? well, pirate 3 will vote no regardless of the offer from 1 or 2 (i blog is right). If pirate 3 decides, pirate 4 will accept one coin to be sure while 3 gets 99. 5 knows this and must prevent pirate 3 from deciding. His offer will not change regardless if pirate 1 or 2 is offering, he still gets one coin. So pirate 1 must offer 4 and 5 the one coin a piece. Here is why 2 would not accept one coin from pirate 1: he knows that 4 and 5 would take the same deal from him that they would take from pirate 1. so he must vote no, along with pirate 3 who always votes no.

In order to be sure that a pirate will vote yes, the pirate must be offered one more coin than he can expect from an offer coming later. Using that logic, I'm going to change my answer.

Please note that pirate 1 is the first pirate to divide the money.

Pirate 3 can give 1 coin to pirate 4 and get his vote, since pirate 4 can never win pirate 5's vote.

So pirate 2 can give 2 coins to pirate 4 (1 more than the minimum he can expect from pirate 3) and 1 coin to pirate 5 (also 1 more than the minimum he can expect from pirate 3) and get their votes, keeping 97 for himself. Pirate 3 gets none if pirate 2 divides the money.

In order to get pirate 5's vote, pirate 1 must offer 2 coins to pirate 5. In order to get pirate 3's vote, he must offer 1 coin to pirate 3.

Pirate 5 is satisfied because he gets 1 more coin than if pirate 2 divides it.

Pirate 3 votes for the division because he wouldn't get any coins if pirate 2 gets to divide the coins.

i blog, i like what you say about pirate 5, but he's getting 1 coin from 2, so why would he demand 1 more from pirate 1? just to be an asshole? maybe you're right and he holds out for two....anyway, three will never vote for yes, he knows 99 are coming his way. four is in the bag with 1.

one last correction...pirate 4 knows he's getting two from pirate 2, and nothing from pirate 1, while 5 is offered 2. Yet pirate 4 determines 1 is better than zero, knowing pirate 5 will be bought off with two. Pirate 4 will accept one coin from pirate 1.

What if we consider pirates 3 and 4 can make deals before the vote, and they both agree to vote no for anything pirates 1 and 2 do, since pirate 3 will tell pirate 4 that he'll give him 1 more than either of the others will?

If pirate 1 can convince pirates 2 and 5 that this is true, pirate 2 would not be able to keep any coins when it is his turn, so he would take 1 coin from pirate 1. Pirate 5 would also take 1 coin, leaving pirate 1 with 98.

Pirate 1 needs pirate 2's vote along with pirate 5's vote in order to get at least half of the votes.

Pirate 1 needs to convince pirate 2 that if pirate 2 gets to divide the money, both pirates 3 and 4 will vote against it, leaving pirate 2 with none and losing his life. So then pirate 1 can offer 1 coin to pirate 2 and he is satisfied, voting for it.

This can work because the most pirate 2 can offer pirate 4 is 98 coins, whereas pirate 3 can tell pirate 4 that he'll give him 1 more than that, 99 coins, to get pirate 4 to vote the way he wants.